As a seasoned tutor on UrbanPro, I can guide you through this physics problem step by step.

Given: Initial velocity of the stone, u = 0 m/s (since it's dropped) Acceleration of the stone, a = g = 9.8 m/s^2 (acceleration due to gravity) Time, t = 11 s

(a) To find the velocity of the stone at t = 11 s, we can use the equation of motion:

v=u+atv=u+at

Substituting the values:

v=0+(9.8×11)v=0+(9.8×11) v=107.8 m/sv=107.8m/s

So, the velocity of the stone at t=11st=11s is 107.8 m/s107.8m/s.

(b) Now, to find the acceleration of the stone at t=11st=11s, we know that the acceleration due to gravity is acting on the stone throughout its motion, so the acceleration remains constant and equal to 9.8 m/s29.8m/s2.

Therefore, the acceleration of the stone at t=11st=11s is 9.8 m/s29.8m/s2.

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As an experienced tutor registered on UrbanPro, I'd be happy to help you with this physics problem.

When the string is cut, the bob will continue its motion with the velocity it had at the instant of cutting the string. Let's analyze both scenarios:

(a) When the bob is at one of its extreme positions: At the extreme position, all of the kinetic energy of the system is converted into potential energy. Therefore, the velocity of the bob at this point is zero. When the string is cut, the bob will simply fall vertically downward due to gravity.

(b) When the bob is at its mean position: At the mean position, the bob has its maximum kinetic energy and zero potential energy. The velocity of the bob at this point is given as 1 m/s. When the string is cut, the bob will continue its motion along a tangent to its trajectory at that point. Since the bob has no initial vertical velocity, it will follow a projectile motion under gravity.

So, in summary: (a) The trajectory is a simple vertical fall. (b) The trajectory is a projectile motion under gravity.

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Sure! Let's solve this problem step by step.

1. Identify the Forces: When the masses are released, the force due to gravity acts on each mass. The heavier mass (12 kg) will experience a greater force downward than the lighter mass (8 kg).

2. Apply Newton's Second Law: The net force on each mass will be the difference between the gravitational force pulling them down and the tension in the string pulling them up. The formula for this is:

   Fnet=m⋅aFnet=m⋅a

   where FnetFnet is the net force, mm is the mass, and aa is the acceleration.

3. Tension in the String: Since the string is light and inextensible, the tension in the string is the same throughout.

4. Acceleration Calculation: We can set up equations for each mass and solve them simultaneously.

   For the 8 kg mass: Fnet=T−mgFnet=T−mg 8a=T−8×9.88a=T−8×9.8 (as g=9.8 m/s2g=9.8m/s2)

   For the 12 kg mass: Fnet=mg−TFnet=mg−T 12a=12×9.8−T12a=12×9.8−T

5. Solve the Equations: Now, we can solve these equations simultaneously to find the acceleration and tension.

   Adding the two equations: 8a+12a=(12×9.8)−(8×9.8)8a+12a=(12×9.8)−(8×9.8) 20a=117.620a=117.6 a=117.620a=20117.6

   Now, plug the value of aa into one of the equations to find the tension:

   8(117.620)=T−8×9.88(20117.6)=T−8×9.8 T=8(117.620)+78.4T=8(20117.6)+78.4

6. Calculate: Let's calculate the values. a=117.620=5.88 m/s2a=20117.6=5.88m/s2 T=8(117.620)+78.4T=8(20117.6)+78.4

So, the acceleration of the masses is 5.88 m/s25.88m/s2, and the tension in the string is TT. You can plug in the value of TT to get the numerical value. This is how you can approach and solve this problem. If you need further assistance or clarification, feel free to ask!

Certainly! When tutoring on UrbanPro, I always strive to provide clear explanations to help students understand complex concepts like this one. Let's break down the scenario.

In the laboratory frame of reference, if a nucleus at rest disintegrates into two smaller nuclei, we need to consider the principles of conservation of momentum and conservation of energy.

1. Conservation of Momentum: Before the disintegration, the total momentum of the system (the nucleus at rest) is zero since the nucleus is stationary. After disintegration, the total momentum must still be zero according to the law of conservation of momentum. If the products move in opposite directions, their momenta will cancel each other out, resulting in a net momentum of zero.

2. Conservation of Energy: The total energy before and after the disintegration must remain constant. Since the nucleus was at rest initially, its energy is purely rest energy (mc^2). After disintegration, this energy is distributed among the kinetic energies of the two smaller nuclei. However, the total kinetic energy must still equal the initial rest energy to comply with the conservation of energy.

Therefore, for the two smaller nuclei to move in opposite directions, they must have equal but opposite momenta and equal kinetic energies. This ensures that both momentum and energy are conserved in the process.

In tutoring sessions on UrbanPro, I often illustrate these concepts with examples and diagrams to make them more accessible to students, helping them grasp the underlying physics principles effectively.

As an experienced tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem.

When two billiard balls collide, they experience a change in momentum due to the impulse imparted during the collision. The impulse experienced by each ball can be calculated using the principle of conservation of momentum.

First, let's define the given values:

• Mass of each ball (m) = 0.05 kg
• Initial speed of each ball (u) = 6 m/s (opposite directions)
• Final speed of each ball after collision (v) = 6 m/s (rebounding with the same speed)

The change in momentum (∆p) of each ball can be calculated using the formula:

∆p = mv - mu

Since the final speed (v) is the same as the initial speed (u) for each ball, the change in momentum (∆p) simplifies to:

∆p = m(v - u)

Substituting the given values:

∆p = 0.05 kg * (6 m/s - (-6 m/s)) ∆p = 0.05 kg * (12 m/s) ∆p = 0.6 kg·m/s

Therefore, the impulse imparted to each ball due to the other is 0.6 kg·m/s. This implies that each ball experiences a change in momentum of 0.6 kg·m/s due to the collision with the other ball. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is a great resource for finding online coaching and tuition services for physics and other subjects.